Is there something wrong with brackets? $f(2x+(f(y)+f(f(y))=4x+8y$ [closed]












0












$begingroup$



$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving







share|cite|improve this question











$endgroup$



closed as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd Jan 6 at 13:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 5




    $begingroup$
    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 11:04












  • $begingroup$
    I think two more parentheses should be added
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 11:14










  • $begingroup$
    It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    $endgroup$
    – ancientmathematician
    Jan 5 at 11:14










  • $begingroup$
    @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    $endgroup$
    – Elementary
    Jan 5 at 11:15


















0












$begingroup$



$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving







share|cite|improve this question











$endgroup$



closed as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd Jan 6 at 13:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 5




    $begingroup$
    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 11:04












  • $begingroup$
    I think two more parentheses should be added
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 11:14










  • $begingroup$
    It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    $endgroup$
    – ancientmathematician
    Jan 5 at 11:14










  • $begingroup$
    @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    $endgroup$
    – Elementary
    Jan 5 at 11:15
















0












0








0





$begingroup$



$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving







share|cite|improve this question











$endgroup$





$ x,y inmathbb{R}$ and $f:mathbb{R} rightarrow mathbb{R}$, find a function that,



$$f(2x+(f(y)+f(f(y))=4x+8y$$



A) $f(x)=2^x$



B) $f(x)=2x$



C) $f(x)=2^x-3$



D) $f(x)=2x^2-3$



E) $f(x)=4x-2$




My problem is,it seems to me that there's something wrong with brackets. Or do I think wrong?






algebra-precalculus contest-math problem-solving




algebra-precalculus contest-math problem-solving






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 11:47







Elementary

















asked Jan 5 at 11:02









ElementaryElementary

361111




361111




closed as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd Jan 6 at 13:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Nosrati, Leucippus, user91500, Paul Frost, Abcd Jan 6 at 13:09


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Leucippus, user91500, Paul Frost, Abcd

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    $begingroup$
    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 11:04












  • $begingroup$
    I think two more parentheses should be added
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 11:14










  • $begingroup$
    It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    $endgroup$
    – ancientmathematician
    Jan 5 at 11:14










  • $begingroup$
    @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    $endgroup$
    – Elementary
    Jan 5 at 11:15
















  • 5




    $begingroup$
    Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 5 at 11:04












  • $begingroup$
    I think two more parentheses should be added
    $endgroup$
    – Mostafa Ayaz
    Jan 5 at 11:14










  • $begingroup$
    It's not unamibiguous, though, is it @MauroALLEGRANZA ?
    $endgroup$
    – ancientmathematician
    Jan 5 at 11:14










  • $begingroup$
    @MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
    $endgroup$
    – Elementary
    Jan 5 at 11:15










5




5




$begingroup$
Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 11:04






$begingroup$
Yes: the number of left ones does not match the number of right ones. But it is enough to add the two missing right parentheses.
$endgroup$
– Mauro ALLEGRANZA
Jan 5 at 11:04














$begingroup$
I think two more parentheses should be added
$endgroup$
– Mostafa Ayaz
Jan 5 at 11:14




$begingroup$
I think two more parentheses should be added
$endgroup$
– Mostafa Ayaz
Jan 5 at 11:14












$begingroup$
It's not unamibiguous, though, is it @MauroALLEGRANZA ?
$endgroup$
– ancientmathematician
Jan 5 at 11:14




$begingroup$
It's not unamibiguous, though, is it @MauroALLEGRANZA ?
$endgroup$
– ancientmathematician
Jan 5 at 11:14












$begingroup$
@MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
$endgroup$
– Elementary
Jan 5 at 11:15






$begingroup$
@MauroALLEGRANZA Doing what you say, are you sure that question can be fixed only in one version?
$endgroup$
– Elementary
Jan 5 at 11:15












1 Answer
1






active

oldest

votes


















4












$begingroup$

Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    $endgroup$
    – Elementary
    Jan 5 at 11:43






  • 1




    $begingroup$
    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    $endgroup$
    – Ennar
    Jan 5 at 11:48










  • $begingroup$
    +1. Perfectly well explained.
    $endgroup$
    – Lucas Henrique
    Jan 5 at 12:40


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    $endgroup$
    – Elementary
    Jan 5 at 11:43






  • 1




    $begingroup$
    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    $endgroup$
    – Ennar
    Jan 5 at 11:48










  • $begingroup$
    +1. Perfectly well explained.
    $endgroup$
    – Lucas Henrique
    Jan 5 at 12:40
















4












$begingroup$

Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    $endgroup$
    – Elementary
    Jan 5 at 11:43






  • 1




    $begingroup$
    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    $endgroup$
    – Ennar
    Jan 5 at 11:48










  • $begingroup$
    +1. Perfectly well explained.
    $endgroup$
    – Lucas Henrique
    Jan 5 at 12:40














4












4








4





$begingroup$

Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.






share|cite|improve this answer









$endgroup$



Looks like the red parenthesis is superfluous: $$f(2x+color{red}{(}f(y)+f(f(y))=4x+8y$$
since $$f(y)+f(f(y)) = color{red}{(}f(y)color{red}{)}+f(f(y)) = color{red}{(}f(y)+f(f(y))color{red}{)},$$
so let us just ignore it.



That leaves us with three cases:




  1. $f(2xcolor{blue}{)}+f(y)+f(f(y))=4x+8y$

  2. $f(2x+f(y)color{blue}{)}+f(f(y))=4x+8y$

  3. $f(2x+f(y)+f(f(y))color{blue}{)}=4x+8y$


I believe the correct one should be 2., since it's the only one that one of the proposed solutions works for.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 5 at 11:35









EnnarEnnar

14.7k32445




14.7k32445








  • 1




    $begingroup$
    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    $endgroup$
    – Elementary
    Jan 5 at 11:43






  • 1




    $begingroup$
    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    $endgroup$
    – Ennar
    Jan 5 at 11:48










  • $begingroup$
    +1. Perfectly well explained.
    $endgroup$
    – Lucas Henrique
    Jan 5 at 12:40














  • 1




    $begingroup$
    (+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
    $endgroup$
    – Elementary
    Jan 5 at 11:43






  • 1




    $begingroup$
    @Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
    $endgroup$
    – Ennar
    Jan 5 at 11:48










  • $begingroup$
    +1. Perfectly well explained.
    $endgroup$
    – Lucas Henrique
    Jan 5 at 12:40








1




1




$begingroup$
(+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
$endgroup$
– Elementary
Jan 5 at 11:43




$begingroup$
(+1) By the way thank you for not saying the answer. I'll solve it myself according to your correction.
$endgroup$
– Elementary
Jan 5 at 11:43




1




1




$begingroup$
@Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
$endgroup$
– Ennar
Jan 5 at 11:48




$begingroup$
@Beginner, you are welcome. If you are doing this for practice, I would still try the other two options as well. It might help you see why some of the proposed solutions cannot work, no matter the parentheses.
$endgroup$
– Ennar
Jan 5 at 11:48












$begingroup$
+1. Perfectly well explained.
$endgroup$
– Lucas Henrique
Jan 5 at 12:40




$begingroup$
+1. Perfectly well explained.
$endgroup$
– Lucas Henrique
Jan 5 at 12:40



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